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Full FOPL is undecidable
sketch of proof by reduction to Halting problems
CS3518
Buchi, J.R. (1962) “Turing machines and the Entscheidungsproblem,”
Boolos, G. & Jeffrey, R. (1989) “Chapter 10: First-Order Logic is Undecidable,” in
their Computability and Logic (Cambridge University Press).
Before we start
Proofs in the literature make use of different types
of Turing Machine (TM)
• 
• 
• 
Sometimes the alphabet is just {blank,1}
(1 is called a mark)
Sometimes transition rules can only either
rewrite (symbol : symbol) or move right or
left: (symbol: R) or (symbol: L)
Sometimes there is no accept or reject state;
halt when no further transition rules apply
Our assumptions
• 
• 
• 
Alphabet = {blank, 1}
Notation: blank = S0 I = S1
Transition rules that can only either
rewrite a (symbol : symbol) or move
right or left: (symbol : R) or (symbol: L)
There are accept and reject states; TM
halts when one of these is reached
We’ve proven: monadic FOPL is decidable
• We’ll see now (in outline) why full FOPL is undecidable
• By inspecting the proof, we’ll see that dyadic FOPL (i.e., no
predicate has more than 2 arguments) is undecidable
Method
• Establish an algorithm for recasting Turing machines as
formulas in FOPL. Call this set of formulas: Delta.
• Establish a formula (ϕ ) saying that a TM halts
• Ensure that a Turing Machine halts on input n iff h
Delta |= phih
• Each deduction step corresponds with a “step” (from t to t+1) in the TM
Constructing an algorithm A
input
output
A
Before we start (logic notation)
For predications, we use infix notation
without brackets:
Argument1 Predicate Argument2
instead of prefix notation with brackets:
Predicate (Argument1, Argument2)
Delta contains 3 kinds of formulas
1. Background theory
Background “Number Logic”
x’ means x+1
Notation
using 3-place predicates Q and S
t Qi x : at time t, state = i and square = x
t Sj x : at t, square x contains symbol Sj
2. A formula representing the input configuration
Initial Configuration
the first n squares on the tape contain S1
all other squares on the tape contain S0
3. Transition rules of the TM
Examples in the next slides
i
Sj : Sk
m
If the machine is in state qi at time t and is then scanning square
number x on which symbol Sj occurs, then at time t+1 the machine
is in state qm scanning square number x, where the symbol Sk
occurs, and in all squares other than x, the same symbols appear at
time t+1 as appeared at time t (for all t and x).
i
Sj : R
m
If the machine is in state qi at time t and is then scanning square
number x on which symbol Sj occurs, then at time t+1 the machine
is in state qm scanning square number x+1, and in all squares the
same symbols appear at time t+1 as appeared at time t (for all t
and x).
i
Sj : L
m
If the machine is in state qi at time t and is then scanning square
number x+1 on which symbol Sj occurs, then at time t+1 the
machine is in state qm scanning square number x, and in all squares
the same symbols appear at time t+1 as appeared at time t (for all
t and x).
halting: ϕh
• The question is: does the TM reach an accept
or reject configuration?
• In other words: is the following a logical
consequence of all the formulas in Delta:
∃x∃t[tQi & (Accept(i) v Reject(i))]
• Call this formula ϕ
h
halting: ϕh
• Another way of putting this:
• The transition rules allow us to prove
formulas of the form: “there exists a time t at
which (…)”
• The TM halts iff we can prove “there exist a
time t at which an accep state is reached
∃x∃t[tQi & (Accept(i) v Reject(i))]
We omit:
• Proof that Delta |= ϕ
TM halts on input n
h
iff Proof by contradiction:
Suppose every question of the form
FOPL Premisses |= FOPL Conclusion
is decidable
It follows (by algorithm A) that the halting
problem is also decidable
Since the halting problem is not decidable, what
we supposed cannot be true
• Once again, this proof uses reduction from
the halting problem
• Other methods exist, but they require
deeper understanding of logic
Bonus: Inspecting this proof
• Observe: None of the formulas in Delta or D
contains predicates with more than 3
arguments
• 
Hence triadic FOPL is undecidable
Inspecting this proof
• Formulas can in fact be simplified so all 3place predicates become 2-place:
•  Q t : at time t, state = i
•  @tx : at time t, reading square x
•  Mtx : at time t, square x is not blank
• So even dyadic FOPL is undecidable
i